# Ambiguous Case Given angle B = 14,\ c = 14.0 and b = 14.1

Ambiguous Case
Given angle $$\displaystyle{B}={14},\ {c}={14.0}$$ and $$\displaystyle{b}={14.1}$$
How many triangles can be expected.
Note that you have to differentiate between right and non-right triangles
Select one:
a. No triangles
b. 1 right triangle
c. 1 non-right triangle
d. 2 possible triangleSK

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Louise Eldridge
Step 1
Sine rule:

$$\displaystyle{\frac{{{\sin{{B}}}}}{{{b}}}}={\frac{{{\sin{{C}}}}}{{{c}}}}$$
$$\displaystyle\Rightarrow{\frac{{{\sin{{\left({14}^{{\circ}}\right)}}}}}{{{14}\cdot{1}}}}={\frac{{{\sin{{c}}}}}{{{14}}}}$$
$$\displaystyle\Rightarrow{\sin{{c}}}={\frac{{{14}\cdot{\sin{{\left({14}^{{\circ}}\right)}}}}}{{{14}\cdot{1}}}}$$
$$\displaystyle\Rightarrow{\sin{{c}}}={0}\cdot{240206137}$$
$$\displaystyle{C}_{{{1}}}={23}\cdot{899}\approx{13}\cdot{9}^{{\circ}}$$
$$\displaystyle{C}_{{{2}}}={180}-{13}\cdot{9}^{{\circ}}\approx{166}\cdot{1}$$ (Which is Not possible)
$$\displaystyle\therefore{B}={14},\ {C}={23}\cdot{9}^{{\circ}},\ {A}={180}-{\left({14}+{13}\cdot{9}\right)}$$
$$\displaystyle{A}={152}\cdot{1}^{{\circ}}$$
Step 2
$$\displaystyle{\frac{{{\sin{{B}}}}}{{{b}}}}={\frac{{{\sin{{A}}}}}{{{a}}}}$$
$$\displaystyle\Rightarrow{\frac{{{\sin{{14}}}}}{{{14}\cdot{1}}}}={\frac{{{\sin{{\left({152}\cdot{2}\right)}}}}}{{{a}}}}\Rightarrow{a}={\frac{{{\sin{{\left({152}\cdot{1}\right)}}}\cdot{14}\cdot{7}}}{{{\sin{{14}}}}}}$$
$$\displaystyle{a}={27}\cdot{27}$$ unit
Only one Triangle (Non-Right Triangle) is possible